The Idea behind z-scores, and Their Relation to Standard Deviation

Date: 06/24/2010 at 20:49:07
From: Charliemagne
Subject: z-scores or z values
I know the definition of a z-score; I know how to compute it, too. But I
don't understand why a z-score measures the number of standard deviations
an observation is above or below the mean.
What is the concept behind z-scores? What connection do they have to
standard deviation?

Date: 06/24/2010 at 21:43:46
From: Doctor Peterson
Subject: Re: z-scores or z values
Hi, Charliemagne.
Texts should always explain what I'm about to say, but I get the
impression that it's not uncommon to treat z-scores as a bit of magic that
you don't need to understand. It's really pretty simple.
The idea is to scale the variable to transform any normal distribution to
a STANDARD normal distribution (with mean 0 and standard deviation 1), so
that we can just make tables for the latter and not need to handle every
possible normal distribution separately.
This scaling involves calling the mean zero, and then measuring the x-axis
using the standard deviation as our unit. That is, the z-score is just
"how many standard deviations x is away from the mean."
Take an example. Suppose the mean of a normal distribution is 100, and the
standard deviation is 15; and suppose our value of x is 130. How many
standard deviations from 100 is 130? We first find how far it is:
130 - 100 = 30
But we want to measure this using the standard deviation, 15. How many
15's are there in 30? We divide 30 by 15 and get the answer, 2. That is,
130 is 2 standard deviations from the mean, 100; so that is what we call
the z-score.
What we just did was to subtract the mean from x, and divide by the
standard deviation:
x - m
-----
s
And that is the definition of the z-score.
Another way to see it is to look at a number line labeled with values
of x:
<---+-------+-------+-------+-------+--->
70 85 100 115 130 x
Subtracting the mean tells how far x is from the mean:
<---+-------+-------+-------+-------+--->
-30 -15 0 15 30 x - m
Dividing by the standard deviation tells how many standard deviations x
is from the mean:
<---+-------+-------+-------+-------+--->
-2 -1 0 1 2 z = (x - m)/s
Since 100 was the mean of x, 0 is the mean of z. And the standard
deviation of z is 1. This makes a standard normal distribution, which you
can look up in the table.
Here's another explanation from our archive:
Calculating and Interpreting Z-Scores
http://mathforum.org/library/drmath/view/68384.html
Does that help?
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/